GENERAL ANALYSIS OF MAXIMA/MINIMA IN CONSTRAINED
OPTIMIZATION PROBLEMS
1. S TATEMENT
OF THE
P ROBLEM
Consider the problem defined by
maximize f (x)
x
subject to g(x) = 0
where g(x) = 0 denotes an m × 1 vector of constraints, m < n. We can also write this as
max
x1 , x2 ,...xn
f (x1 , x2 , . . . , xn )
subject to
g1(x1 , x2 , . . . , xn ) = 0
g2(x1 , x2, . . . , xn ) = 0
..
.
(1)
gm (x1 , x2, . . . , xn ) = 0
The solution can be obtained using the Lagrangian function
L(x; λ) = f (x) − λ0g(x) where λ0 = (λ1, λ2, . . . , λm)
= f (x1, x2 , . . . ) − λ1g1(x) − λ2g2 (x) − · · · − λm gm (x)
(2)
Notice that the gradient of L will involve a set of derivatives, i.e.
∂g
∇xL = ∇x f (x) −
λ
∂x
where

∂g1(x∗)
 ∂x1



 ∂g1(x∗)

 ∂x2
∂g
= Jg = 

∂x

..

.



 ∂g1(x∗)
∂xn
Date: October 7, 2004.
1
∂g2(x∗)
∂x1
...
∂g2(x∗)
∂x2
...
..
.
..
.
∂g2(x∗)
∂xn
...

∂gm (x∗)
∂x1 



∂gm (x∗) 

∂x2 



..

.



∗
∂gm (x ) 
∂xn
(3)
2
GENERAL ANALYSIS OF MAXIMA/MINIMA IN CONSTRAINED OPTIMIZATION PROBLEMS
There will be one equation for each x. There will also be equations involving the derivatives of L with respect to each λ.
2. N ECESSARY C ONDITIONS
FOR AN
E XTREME P OINT
The necessary conditions for an extremum of f with the equality constraints g(x) = 0
are that
∇L(x∗, λ∗) = 0
(4)
where it is implicit that the gradient in (3) is with respect to both x and λ.
3. S UFFICIENT C ONDITIONS
FOR AN
E XTREME P OINT
3.1. Statement of Conditions. Let f, g1, . . . , gm be twice continuously differentiable realvalued functions on Rn . If there exist vectors x∗ Rn , λ∗ Rm such that
∇L(x∗, λ∗) = 0
(5)
and for every non-zero vector z Rn satisfying
z 0∇gi (x∗) = 0,
i = 1, . . . , m
(6)
it follows that
z 0 ∇2x L(x∗, λ∗)z > 0,
(7)
then f has a strict local minimum at x∗ , subject to gi (x) = 0, i = 1, . . . , m. If the inequality in (7) is reversed, then f has strict local maximum at x∗ . The idea is that if equation 5
holds, then if equation 7 holds for all vectors satisfying equation 6, f will have a strict local
minimum at x∗.
3.2. Checking the Sufficient Conditions. These conditions for a maximum or minimum
can be stated in terms of the Hessian of the Lagrangian function (or bordered Hessian).
Let f, g1, . . . , gm be twice continuously differentiable real valued functions. If there exist
vectors x∗ Rn , λ∗ Rm , such that
∇L(x∗, λ∗) = 0
and if
(8)
GENERAL ANALYSIS OF MAXIMA/MINIMA IN CONSTRAINED OPTIMIZATION PROBLEMS

∂ 2L(x∗, λ∗)
 ∂x ∂x
1
1


·


·


 2 ·∗ ∗
 ∂ L(x , λ )

 ∂x ∂x
p
1

m
(−1) det 

 ∂g1(x∗ )


∂x1


·


·


·

 ∂gm(x∗ )
∂x1
...
·
·
·
...
...
·
·
·
...
∂ 2L(x∗, λ∗)
∂x1 ∂xp
·
·
·
∂ 2L(x∗, λ∗)
∂xp ∂xp
∂g1 (x∗)
∂x1
·
·
·
∂g1 (x∗)
∂xp
∂g1(x∗ )
∂xp
·
·
·
∂gm (x∗)
∂xp
...
·
·
·
...
0
...
·
·
·
·
·
·
0
...

∂gm (x∗)
∂x1 


·


·


·

∗ 
∂gm(x ) 
d
∂xp 

>0



0



·


·


·


0
3
(9)
for p = m + 1, . . . , n, then f has a strict local minimum at x∗ , such that
gi (x∗) = 0, i = 1, . . . , m.
(10)
We check the determinants in (9) starting with the one that has m + 1 elements in each
row and column of the Hessian and m+1 elements in each row or column of the derivative
of a given constraint with respect to x. Note that m does not change as we check the various
determinants so that they will all be of the same sign for a given m.
If there exist vectors x∗ Rn , λ∗ Rm , such that
∇L(x∗ , λ∗ ) = 0
(11)
and if

∂ 2L(x∗, λ∗)
 ∂x ∂x
1
1


·


·


 2 ·∗ ∗
 ∂ L(x , λ )

 ∂x ∂x
p
1

(−1)p det 

 ∂g1(x∗ )


∂x1


·


·


·

 ∂gm(x∗ )
∂x1
...
·
·
·
...
∂ 2L(x∗, λ∗)
∂x1∂xp
·
·
·
∂ 2L(x∗, λ∗)
∂xp ∂xp
∂g1(x∗)
∂x1
·
·
·
∂g1(x∗)
∂xp
...
·
·
·
...
∗
...
·
·
·
...
∂g1(x )
∂xp
·
·
·
∂gm(x∗ )
∂xp
0
...
·
·
·
·
·
·
0
...

∂gm (x∗)
∂x1 


·


·


·

∗
∂gm (x ) 

∂xp 

>0



0



·


·


·


0
(12)
4
GENERAL ANALYSIS OF MAXIMA/MINIMA IN CONSTRAINED OPTIMIZATION PROBLEMS
for p = m + 1, . . ., n then f has a strict local maximum at x∗, such that
gi (x∗) = 0,
(13)
i = 1, . . ., m.
We check the determinants in (12) starting with the one that has m + 1 elements in
each row and column of the Hessian and m + 1 elements in each row or column of the
derivative of a given constraint with respect to x. Note that p changes as we check the
various determinants so that they will alternate in sign for a given m.
Consider the case where n = 2 and m = 1. Note that the first matrix we check has
p = m + 1 = 2. Then the condition for a minimum is

∂ 2L(x∗ , λ∗ )
 ∂x1∂x1


 2
 ∂ L(x∗ , λ∗ )
(−1) det 
 ∂x2∂x1


 ∂g(x∗)
∂x1

∂g(x∗)
∂x1 



∂g(x∗) 
>0
∂x2 



0
∂ 2L(x∗, λ∗)
∂x1∂x2
∂ 2L(x∗, λ∗)
∂x2∂x2
∂g(x∗)
∂x2
(14)
This, of course, implies

∂ 2 L(x∗, λ∗)
 ∂x1∂x1


 2
 ∂ L(x∗, λ∗)
det 
 ∂x2∂x1


 ∂g(x∗)
∂x1
∂ 2L(x∗ , λ∗ )
∂x1∂x2
∂ 2L(x∗ , λ∗ )
∂x2∂x2
∂g(x∗)
∂x2

∂g(x∗)
∂x1 



∂g(x∗) 
<0
∂x2 



0
(15)
The condition for a maximum is

∂ 2 L(x∗, λ∗)
 ∂x1 ∂x1


 2
 ∂ L(x∗, λ∗)
(−1)2 det 
 ∂x2 ∂x1


 ∂g(x∗)
∂x1
This, of course, implies
∂ 2 L(x∗, λ∗)
∂x1∂x2
∂ 2 L(x∗, λ∗)
∂x2∂x2
∂g(x∗)
∂x2

∂g(x∗)
∂x1 


∗ 
∂g(x ) 
>0
∂x2 



0
(16)
GENERAL ANALYSIS OF MAXIMA/MINIMA IN CONSTRAINED OPTIMIZATION PROBLEMS

∂ 2 L(x∗, λ∗)
 ∂x1 ∂x1


 2
 ∂ L(x∗, λ∗)
det 
 ∂x2 ∂x1


 ∂g(x∗)
∂x1
∂ 2 L(x∗, λ∗)
∂x1∂x2
∂ 2 L(x∗, λ∗)
∂x2∂x2
∂g(x∗)
∂x2

∂g(x∗)
∂x1 



∂g(x∗) 
>0
∂x2 



0
5
(17)
Also consider the case where n = 3 and m = 1. We start with p = m + 1 = 2 and
continue until p = n. Then the condition for a minimum is

∂ 2L(x∗, λ∗)
 ∂x1∂x1


 2
 ∂ L(x∗, λ∗)
(−1) det 
 ∂x2∂x1


 ∂g(x∗)
∂x1

∂ 2L(x∗ , λ∗ )
 ∂x1∂x1


 2
 ∂ L(x∗ , λ∗ )

 ∂x2∂x1

(−1) det 
 2
 ∂ L(x∗ , λ∗ )

 ∂x3∂x1


 ∂g(x∗)
∂x1
The condition for a maximum is
∂ 2L(x∗, λ∗)
∂x1∂x2
∂ 2L(x∗ , λ∗)
∂x2∂x2
∂g(x∗)
∂x2
∂ 2L(x∗, λ∗)
∂x1∂x2
∂ 2L(x∗, λ∗)
∂x1 ∂x3
∂ 2L(x∗, λ∗)
∂x2∂x2
∂ 2L(x∗, λ∗)
∂x2 ∂x3
∂ 2L(x∗, λ∗)
∂x3∂x2
∂ 2L(x∗, λ∗)
∂x3 ∂x3
∂g(x∗)
∂x2
∂g(x∗)
∂x3

∂g(x∗)
∂x1 



∂g(x∗) 
>0
∂x2 



0

∂g(x∗)
∂x1 



∂g(x∗) 

∂x2 

>0
∗ 
∂g(x ) 

∂x3 



0
(18)
6
GENERAL ANALYSIS OF MAXIMA/MINIMA IN CONSTRAINED OPTIMIZATION PROBLEMS

∂ 2L(x∗, λ∗)
 ∂x1∂x1


 2
 ∂ L(x∗, λ∗)
(−1)2 det 
 ∂x2∂x1


 ∂g(x∗)
∂x1

∂ 2L(x∗ , λ∗ )
 ∂x1∂x1


 2
 ∂ L(x∗ , λ∗ )

 ∂x2∂x1

3
(−1) det 
 2
 ∂ L(x∗ , λ∗ )

 ∂x1∂x1


 ∂g(x∗)
∂x1
∂ 2L(x∗, λ∗)
∂x1 ∂x2
∂ 2L(x∗, λ∗)
∂x2 ∂x2
∂g(x∗)
∂x2
∂ 2L(x∗, λ∗)
∂x1∂x2
∂ 2L(x∗, λ∗)
∂x1 ∂x3
∂ 2L(x∗, λ∗)
∂x2∂x2
∂ 2L(x∗, λ∗)
∂x2 ∂x3
∂ 2L(x∗, λ∗)
∂x1∂x2
∂ 2L(x∗, λ∗)
∂x1 ∂x3
∂g(x∗)
∂x2
∂g(x∗)
∂x3

∂g(x∗)
∂x1 



∗
∂g(x ) 
>0
∂x2 



0

∂g(x∗)
∂x1 



∂g(x∗) 

∂x2 

>0

∂g(x∗) 

∂x3 



0
(19)
3.3. Sufficient Condition for a Maximum and Minimum and Positive and Negative Definite Quadratic Forms. Note that at the optimum, equation 6 is just linear in the sense that
the derivatives
∂gi(x∗ )
∂xj
∗
are fixed numbers at the point x and we can write equation 6 as
z 0 Jg = 0


∂g1(x∗) ∂g2 (x∗)
∂gm (x∗)
...
 ∂x1
∂x1
∂x1 

  


0

∗
∗
∗ 
∂gm (x )   
 ∂g1(x ) ∂g2 (x )

 0
...
∂x2
∂x2
∂x2  =  . 
(20)
(z1 z2 . . . zn ) 

 .
.




..
..
..
..
0


.
.
.
.


∗
∗
∗
 ∂g1(x ) ∂g2 (x )
∂gm (x ) 
...
∂xn
∂xn
∂xn
o
n
∗
∂gi (x )
where Jg is the matrix
and where there is a column of the Jg for each constraint
∂xj
and a row for each x variable we are considering. This then implies that the sufficient condition for a strict local maximum of the function f is that |HB | has the same sign as (−1)p,
that is the last n − m leading principal minors of HB alternate in sign on the constraint set
GENERAL ANALYSIS OF MAXIMA/MINIMA IN CONSTRAINED OPTIMIZATION PROBLEMS
7
denoted by equation 6. This is the same as the condition that the quadratic form z 0HB z be
negative definite on the constraint set
(21)
z 0 ∇gi(x∗ ) = 0, i = 1, . . ., m
If |HB | and these last n − m leading principal minors all have the same sign as (−1)m ,
then z 0 HB z is positive definite on the constraint set z 0 ∇gi(x∗ ) = 0, i = 1, . . ., m and the
function has strict local minimum at the point x∗ .
If both of conditions are violated by non-zero leading principal minors, then z 0 HB z
is indefinite on the constraint set and we cannot determine whether the function has a
maximum or a minimum.
3.4. Example 1: Minimizing Cost Subject to an Output Constraint. Consider a production function given by
y = 20x1 − x21 + 15x2 − x22
(22)
Let the prices of x1 and x2 be 10 and 5 respectively with an output constraint of 55.
Then to minimize the cost of producing 55 units of output given this prices we set up the
following Lagrangian
L = 10x1 + 5x2 − λ(20x1 − x21 + 15x2 − x22 − 55)
∂L
= 10 − λ(20 − 2x1) = 0
∂x1
(23)
∂L
= 5 − λ(15 − 2x2) = 0
∂x2
∂L
= (−1)(20x1 − x21 + 15x2 − x22 − 55) = 0
∂λ
If we take the ratio of the first two first order conditions we obtain
10
5
⇒ 30 − 4x2
⇒ 10 − 4x2
⇒ x1
20 − 2x1
15 − 2x2
= 20 − 2x1
= −2x1
= 2x2 − 5
=2=
(24)
Now plug this into the negative of the last first order condition to obtain
20(2x2 − 5) − (2x2 − 5)2 + 15x2 − x22 − 55 = 0
Multiplying out and solving for x2 will give
(25)
8
GENERAL ANALYSIS OF MAXIMA/MINIMA IN CONSTRAINED OPTIMIZATION PROBLEMS
40x2 − 100 − (4x22 − 20x2 + 25) + 15x2 − x22 − 55 = 0
⇒ 40x2 − 100 − 4x22 + 20x2 − 25 + 15x2 − x22 − 55 = 0
⇒ −5x22 + 75x2 − 180 = 0
⇒
5x22
⇒
(26)
− 75x2 + 180 = 0
x22
− 15x2 + 36 = 0
Now solve this quadratic equation for x2 as follows
x2 =
15 ±
p
225 − 4(36)
2
√
15 ± 81
=
2
(27)
= 12 or 3
Therefore,
x1 = 2x2 − 5
= 19 or 1
(28)
The Lagrangian multiplier λ can be obtained by solving the first equation that was obtained by differentiating L with respect to x1
10 − λ(20 − (19)) = 0
⇒λ=−
5
9
(29)
10 − λ(20 − 2(1)) = 0
5
⇒λ=
9
To check for a maximum or minimum we set up the bordered Hessian as in
equations 14–17. The bordered Hessian in this case is

∂ 2L(x∗, λ∗ ) ∂ 2L(x∗, λ∗) ∂g(x∗)
 ∂x1∂x1
∂x1 ∂x2
∂x1 





 2
 ∂ L(x∗, λ∗ ) ∂ 2L(x∗, λ∗) ∂g(x∗) 
(30)
HB = 


 ∂x2∂x1
∂x
∂x
∂x
2
2
2





 ∂g(x∗)
∗
∂g(x )
0
∂x1
∂x2
We only need to compute one determinant. We compute the various elements of the
bordered Hessian as follows

GENERAL ANALYSIS OF MAXIMA/MINIMA IN CONSTRAINED OPTIMIZATION PROBLEMS
9
L = 10x1 + 5x2 − λ(20x1 − x21 + 15x2 − x22 − 55)
∂L
= 10 − λ(20 − 2x1)
∂x1
∂L
= 5 − λ(15 − 2x2 )
∂x2
∂ 2L
= 2λ
∂x1 ∂x1
(31)
∂ 2L
=0
∂x1 ∂x2
∂ 2L
= 2λ
∂x2 ∂x2
∂g
= 20 − 2x1
∂x1
∂g
= 15 − 2x2
∂x2
Consider first the point (19, 12, -5/9). The bordered Hessian is given by



HB = 


2λ
0
0
2λ
20 − 2x1 15 − 2x2
x1 = 19,
x2 = 12,
 10
−
 9


HB = 
 0


−18
0
−
10
9
−9
The determinant of the bordered Hessian is
λ=−

−18




−9 


0
20 − 2x1



15 − 2x2


0
5
9
(32)
10
GENERAL ANALYSIS OF MAXIMA/MINIMA IN CONSTRAINED OPTIMIZATION PROBLEMS
10
−
− 10 −9
9
10 9
2
3
|HB | = (−1) −
+ (−1) (0) 9 −9
−9
0
10
= −
(−81) + 0 + (−18)(−20)
9
10 0
−
−9
9
+ (−1)4(−18) −18 −9 0
(33)
= 90 + 360 = 450
Here p = 2 so the condition for a maximum is that (−1)2|HB | > 0, so this point is a
relative maximum.
Now consider the other point, (1, 3, 5/9). The bordered Hessian is given by



HB = 


2λ
0
0
2λ
20 − 2x1
20 − 2x1 15 − 2x2
x1 = 1,
x2 = 3,
 10
0
9


HB = 
0


10
9
18
9
λ=
18



15 − 2x2 


0
5
9
(34)





9


0
The determinant of the bordered Hessian is
|HB | = (−1)2
=
10
9
10
9
9
10
9
9
+ (−1)3 (0)| 9
0
0
9
+ (−1)4(18) 18
0
10
(−81) + 0 + (18)(−20)
9
10 9
9
(35)
= −90 − 360 = −450
The condition for a minimum is that (−1)|HB | > 0, so this point is a relative minimum.
The minimum cost is obtained by substituting into the cost expression to obtain
C = 10(1) + 5(3) = 25
(36)
GENERAL ANALYSIS OF MAXIMA/MINIMA IN CONSTRAINED OPTIMIZATION PROBLEMS
11
3.5. Example 2: Maximizing Output Subject to a Cost Constraint. Consider a production
function given by
y = 30x1 + 12x2 − x21 + x1 x2 − x22
(37)
Let the prices of x1 and x2 be 10 and 4 respectively with an cost constraint of $260.
Then to maximize output with a cost of $260 given these prices we set up the following
Lagrangian
L = 30x1 + 12x2 − x21 + x1 x2 − x22 − λ(10x1 + 4x2 − 260)
∂L
= 30 − 2x1 + x2 − 10λ = 0
∂x1
∂L
= 12 + x1 − 2x2 − 4λ = 0
∂x2
(38)
∂L
= −10x1 − 4x2 + 260 = 0
∂λ
If we take the ratio of the first two first order conditions we obtain
10
30 − 2x1 + x2
= 2.5 =
4
12 + x1 − 2x2
⇒ 30 + 2.5x1 − 5x2 = 30 − 2x1 + x2
(39)
⇒ 4.5x1 = 6x2
⇒ x1 = 1.33̄x2
Now plug this value for x1 into the negative of the last first order condition to obtain
10x1 + 4x2 − 260 = 0
⇒ (10)(1.33̄x2) + 4x2 − 260 = 0
⇒ 13.33̄x2 + 4x2 = 260
⇒ 17.33̄x2 = 260
⇒ x2 = 15
4
⇒ x1 = 7
(15) = 20
3
We can also find the maximum y by substituting in for x1 and x2.
(40)
12
GENERAL ANALYSIS OF MAXIMA/MINIMA IN CONSTRAINED OPTIMIZATION PROBLEMS
y = 30x1 + 12x2 − x21 + x1x2 − x22
= (30)(20) + (12)(15) − (20)2 − (20)(15) − (15)2
= 600 + 180 − 400 + 300 − 225
= 455
(41)
The Lagrangian multiplier λ can be obtained by solving the first equation that was obtained by differentiating L with respect to x1
30 − 2x1 + x2 − 10λ = 0
⇒ 30 − 2(20) + (15) − 10λ = 0
⇒ 30 − 40 + 15 − 10λ = 0
⇒ 5 = 10λ
1
⇒λ=
2
(42)
To check for a maximum or minimum we set up the bordered Hessian as in equations 14–17 where p = 2 and m = 1. The bordered Hessian in this case is

∂ 2L(x∗, λ∗ )
 ∂x1∂x1


 2
 ∂ L(x∗, λ∗ )
HB = 
 ∂x2∂x1


 ∂g(x∗)
∂x1
∂ 2L(x∗, λ∗)
∂x1 ∂x2
∂ 2L(x∗, λ∗)
∂x2 ∂x2
∂g(x∗)
∂x2

∂g(x∗)
∂x1 



∂g(x∗) 

∂x2 



0
We compute the various elements of the bordered Hessian as follows
(43)
GENERAL ANALYSIS OF MAXIMA/MINIMA IN CONSTRAINED OPTIMIZATION PROBLEMS
13
L = 30x1 + 12x2 − x21 + x1x2 − x22 − λ(10x1 + 4x2 − 260)
∂L
= 30 − 2x1 + x2 − 10λ
∂x1
∂L
= 12 + x1 − 2x2 − 4λ
∂x2
∂ 2L
= −2
∂x1∂x1
(44)
∂ 2L
=1
∂x1∂x2
∂ 2L
= −2
∂x2∂x2
∂g
= 10
∂x1
∂g
=4
∂x2
The derivatives are all constants. The bordered Hessian is given by

−2

HB =  1
1
−2
10 4
The determinant of the bordered Hessian is
10


4
(45)
0
−2 4
1 4
1 −2
3
4
|HB| = (−1) (−2) + (−1) (1) + (−1) (10)| 4 0
10 0
10 4 2
= (−2)(−16) − (−40) + (10)(24)
(46)
= 32 + 40 + 240 = 312
The condition for a maximum is that (−1)2|HB | > 0, so this point is a relative maximum.
3.6. Example 3: Maximizing Utility Subject to an Income Constraint. Consider a utility
function given by
1 α2
u = xα
1 x2
Now maximize this function subject to the constraint that
w1 x1 + w2x2 = c0
14
GENERAL ANALYSIS OF MAXIMA/MINIMA IN CONSTRAINED OPTIMIZATION PROBLEMS
Set up the Lagrangian problem:
1 α2
L = xα
1 x2 − λ[w1 x1 + w2 x2 − c0 ]
The first order conditions are
∂L
1 −1 α2
= α 1 xα
x2 − λ w 1 = 0
1
∂x1
∂L
1 α2 −1
= α 2 xα
− λw2 = 0
1 x2
∂x2
∂L
= −w1 x1 − w2 x2 + c0= 0
∂λ
Taking the ratio of the 1st and 2nd equations we obtain
w1
α 1 x2
=
w2
α2 x1
We can now solve the equation for the 2nd quantity as a function of the 1st input quantity
and the prices. Doing so we obtain
x2 =
α 2 x1 w1
α1 w2
Now substituting in the income equation we obtain
w 1 x1 + w 2 x2 = c 0
⇒ w 1 x1 + w 2
α 2 x1 w1
= c0
α1 w2
α 2 w1 w2
x1 = c0
⇒ w 1 x1 +
α1 w2
α 2 w1
x1 = c0
⇒ w 1 x1 +
α1
⇒ x1
α 2 w1
w1 +
= c0
α1
GENERAL ANALYSIS OF MAXIMA/MINIMA IN CONSTRAINED OPTIMIZATION PROBLEMS
15
α2
= c0
⇒ x 1 w1 1 +
α1
⇒ x 1 w1
α1 + α2
= c0
α1
α1
c0
⇒ x1 =
w1 α1 + α 2
We can now get x2 by substitution
x 2 = x1
α 2 w1
α1 w2
α1
α 2 w1
c0
=
w1 α1 + α 2 α1 w2
α2
c0
=
w2 α1 + α 2
We can find the value of the optimal u by substitution
1 α2
u = xα
1 x2
=
α1 α2
α1
c0
α2
c0
w1 α1 + α 2
w2 α1 + α 2
−α1 −α2
1 +α2
1 α2
= cα
w1−α1 w2−α2 αα
0
1 α2 (α2 + α2 )
This can also be written
1 α2
u = xα
1 x2
c0
=
w1
=
α1
α1 + α 2
α1
α1 + α 2
α1 α1 c0
w2
α2
α1 + α 2
α2
α1 + α 2
α2 c0
w1
α2
α1 c0
w2
α2
For future reference note that the derivative of the optimal u with respect to c0 is given
by
16
GENERAL ANALYSIS OF MAXIMA/MINIMA IN CONSTRAINED OPTIMIZATION PROBLEMS
−α1 −α2
1 +α2
1 α2
u = cα
w1−α1 w2−α2 αα
0
1 α2 (α2 + α2 )
∂u
−α1 −α2
1 +α2 −1 −α1
1 α2
= (α1 + α2)cα
w1 w2−α2 αα
0
1 α1 (α2 + α2 )
∂c0
1−α1 −α2
1 +α2 −1 −α1
1 α2
= cα
w1 w2−α2 αα
0
1 α1 (α2 + α2 )
We obtain λ by substituting in either the first or second equation as follows
1 −1 α2
x2 − λw1 = 0
α 1 xα
1
1 −1 α2
α 1 xα
x2
1
w1
1 α2 −1
α2 xα
− λw2 = 0
1 x2
⇒λ=
⇒λ=
1 α2 −1
α 2 xα
1 x2
w2
If we now substitute for x1 and x2 , we obtain
1 −1 α2
α 1 xα
x2
1
w1
α1
c0
x1 =
w1 α1 + α 2
λ=
α2
c0
x2 =
w2 α1 + α 2
α1
⇒λ=
=
α1 −1 α2
c0
α1
c0
α2
w1 α1 + α 2
w2 α1 + α 2
w1
1 +α2 −1 1−α1
1 −1 α2
w1
w2−α2 αα
α2 (α1 + α2 )1−α1 −α2
α 1 cα
0
1
w1
1−α1 −α2
1 +α2 −1 −α1
1 α2
= cα
w1 w2−α2 αα
0
1 α2 (α1 + α2 )
Thus λ is equal to the derivative of the optimal u with respect to c0.
To check for a maximum or minimum we set up the bordered Hessian as in equations 14–17 where p = 2 and m = 1. The bordered Hessian in this case is
GENERAL ANALYSIS OF MAXIMA/MINIMA IN CONSTRAINED OPTIMIZATION PROBLEMS

∂ 2L(x∗ , λ∗ )
 ∂x1∂x1


 2
 ∂ L(x∗ , λ∗ )
HB = 
 ∂x2∂x1


 ∂g(x∗)
∂x1

∂g(x∗)
∂x1 


∗ 
∂g(x ) 

∂x2 



0
∂ 2L(x∗ , λ∗ )
∂x1∂x2
∂ 2L(x∗ , λ∗ )
∂x2∂x2
∂g(x∗)
∂x2
17
(47)
We need compute the various elements of the bordered Hessian as follows
1 α2
L = xα
1 x2 − λ[w1 x1 + w2 x2 − c0 ]
∂L
1 −1 α2
= α 1 xα
x2 − λw1
1
∂x1
∂L
1 α2 −1
= α 2 xα
− λw2
1 x2
∂x2
∂ 2L
1 −2 α2
= (α1)(α1 − 1)xα
x2
1
∂x21
∂ 2L
1 −1 α2 −1
= α 1 α2 xα
x2
1
∂x1 ∂x2
∂ 2L
1 α2 −2
= (α2)(α2 − 1)xα
1 x2
∂x22
∂g
= w1
∂x1
∂g
= w2
∂x2
The derivatives of the constraints are constants. The bordered Hessian is given by



HB = 


1 −2 α2
x2
(α1)(α1 − 1)xα
1
1 −1 α2 −1
α1 α2 xα
x2
1
w1
1 −1 α2 −1
α1 α2 xα
x2
1
(α2 )(α2 −
1 α2 −2
1)xα
1 x2
w2
w1



w2 


0
(48)
To find the determinant of the bordered Hessian, expand by the third row as follows
18
GENERAL ANALYSIS OF MAXIMA/MINIMA IN CONSTRAINED OPTIMIZATION PROBLEMS
(α1)(α1 − 1)xα1−2 xα2 w1
α1α2 xα1 −1 xα2 −1
w1
1
2
1
2
5
|HB | = (−1) w1| + (−1) w2 +0
α1 α2xα1 −1 xα2−1
(α2)(α2 − 1)xα1 xα2 −2 w2
w2 1
2
1 2
4
α1 α2xα1 −1 xα2 −1
(α1)(α1 − 1)xα1−2 xα2 w1
w1
1
2
1
2
= w1 − w2 α
α
−2
α
−1
α
−1
1
2
1
2
(α2)(α2 − 1)x x
α1 α2 x
w2 x2
w2 1 2
1
1 −1 α2 −1
1 α2 −2
= w 1 w2 α1 α2 xα
x2
− w12(α2 )(α2 − 1)xα
1
1 x2
1 −2 α2
1 −1 α2−1
− w22 (α1)(α1 − 1)xα
x2 + w 1 w2 α1 α2 xα
x2
1
1
1 −1 α2 −1
1 α2 −2
1 −2 α2
= 2w1w2α1 α2xα
x2
− w12(α2 )(α2 − 1)xα
− w22 (α1)(α1 − 1)xα
x2
1
1
1 x2
(49)
For a maximum we want this expression to be positive. Rewriting it we obtain
1 −1 α2 −1
1 α2 −2
1 −2 α2
2w1w2α1 α2 xα
x2
− w12(α2 )(α2 − 1)xα
− w22 (α1)(α1 − 1)xα
x2 > 0
1
1
1 x2
(50)
We can also write it in the following convenient way
1 −1 α2 −1
2w1w2α1 α2 xα
x2
1
1 α2 −2
1 α2 −2
+α2 w12xα
− α22 w12xα
1 x2
1 x2
(51)
1 −2 α2
1 −2 α2
+α1 w22 xα
x2 − α21 w22 xα
x2 > 0
1
1
To eliminate the prices we can substitute from the first-order conditions.
w1 =
1 −1 α2
x2
α 1 xα
1
λ
w2 =
1 α2 −1
α 2 xα
1 x2
λ
This then gives
!
2
1 α2 −1
α1 xα1−1
xα
α 2 xα
1
2
1 x2
1 −1 α2 −1
2
x2
α 1 α2 xα
1
λ
λ
!2
!
α1 −1 α2 2
1 −1 α2
α 1 xα
α
x
x
x
1 1
1
2
2
1 α2 −2
1 α2 −2
xα
− α22
xα
1 x2
1 x2
λ
λ
+α2
+α1
1 α2 −1
α 2 xα
1 x2
λ
!2
1 −2 α2
xα
x2 − α21
1
1 α2 −1
α 2 xα
1 x2
λ
!2
1 −2 α2
xα
x2 > 0
1
(52)
GENERAL ANALYSIS OF MAXIMA/MINIMA IN CONSTRAINED OPTIMIZATION PROBLEMS
19
Multiply both sides by λ2 and combine terms to obtain
1 −2 3α2 −2
2α21α22 x3α
x2
1
1 −2 3α2 −2
2 −2
+α21 α2 x3α
x2
− α22 α21 x3α1−2
x3α
1
2
1
(53)
1 −2 3α2 −2
1 −2 3α2 −2
+α1 α22 x3α
x2
− α21 α22 x3α
x2
>0
1
1
1 −2 3α2 −2
Now factor out x3α
x2
to obtain
1
1 −2 3α2 −2
x3α
2α21α22 + α21 α2 − α22 α21 + α1 α22 − α21 α22 >0
x2
1
1 −2 3α2 −2
⇒ x3α
α21 α2 + α1 α22 >0
x2
1
(54)
With positive values for x1 and x2 the whole expression will be positive if the last term
in parentheses is positive. Then rewrite this expression as
α21 α2 + α1 α22 > 0
Now divide both sides by
α21α22
(which is positive) to obtain
1
1
>0
+
α2 α1
(55)
(56)
3.7. Some More Example Problems.
(i) opt [x1x2 ] s. t.
x1 , x2
x1 + x2 = 6
(ii) opt [x1x2 + 2x1 ] s.t.
x1 , x2
4x1 + 2x2 = 60
(iii) opt [x21 + x22 ] s.t
x1 , x2
x1 + 2x2 = 20
(iv) opt [x1x2 ] s.t.
x1 , x2
x21 + 4x22 = 1
1
1
(v) opt [x14 x22 ] s.t.
x1 , x2
2x1 + 8x2 = 60
4. T HE I MPLICIT F UNCTION T HEOREM
4.1. Statement of Theorem. We are often interested in solving implicit systems of equations for m variables, say x1 , x2 , . . . , xm in terms of m+p variables where there are a minimum of m equations in the system. We typically label the variables xm+1 , xm+2 , . . . , xm+p ,
∂xi
y1 , y2 , . . . , yp . We are frequently interested in the derivatives ∂x
where it is implicit that
j
all other xk and all y` are held constant. The conditions guaranteeing that we can solve for
20
GENERAL ANALYSIS OF MAXIMA/MINIMA IN CONSTRAINED OPTIMIZATION PROBLEMS
m of the variables in terms of p variables along with a formula for computing derivatives
is given by the implicit function theorem.
Theorem 1 (Implicit Function Theorem). Suppose that φi are real-valued functions defined on
a domain D and continuously differentiable on an open set D1 ⊂ D ⊂ Rm+p , where p > 0 and
φ(x01, x02 , . . . , x0m , y10, y20 , . . . , yp0 ) = φi (x0 , y 0 ) = 0,
i = 1, 2, . . . , m, and (x0 , y 0 ) ∈ D1 .
0
(57)
0
,y )
Assume the Jacobian matrix [ ∂φi(x
] has rank m. Then there exists a neighborhood Nδ (x0,
∂xj
y0) ⊂ D1 , an open set D2 ⊂ Rp containing y0 and real valued functions ψk , k = 1, 2, . . . , m,
continuously differentiable on D2 , such that the following conditions are satisfied:
x0k = ψk (y 0), k = 1, 2, . . . , m.
(58)
For every y ∈ D2 , we have
φi (ψ1(y), ψ2(y), . . . , ψm (y), y1 , y2 , . . . , yp ) ≡ 0,
i = 1, 2, . . . , m.
or
φi (ψ(y), y) ≡ 0,
(59)
i = 1, 2, . . . , m.
∂φ (x, y)
i
We also have that for all (x,y) ∈ Nδ (x0 , y0 ), the Jacobian matrix [ ∂x
] has rank m. Furtherj
2
more for y ∈ D , the partial derivatives of ψ(y) are the solutions of the set of linear equations
m
X
∂φi (ψ(y), y) ∂ψk (y)
−∂φi (ψ(y), y
=
∂xk
∂yj
∂yj
i = 1, 2, . . . , m
(60)
k=1
4.2. Example with one equation and three variables. Consider one implicit equation
with three variables.
φ(x01 , x02 , y 0 ) = 0
x02
(61)
The implicit function theorem says that we can solve equation 61 for x01 as a function of
and y0 , i.e.,
x01 = ψ1 (x02, y 0 )
(62)
φ(ψ1(x2, y), x2, y) = 0
(63)
and that
The theorem then says that
GENERAL ANALYSIS OF MAXIMA/MINIMA IN CONSTRAINED OPTIMIZATION PROBLEMS
21
∂φ(ψ1(x2 , y), x2, y) ∂ψ1
−∂φ(ψ1(x2, y), x2 , y)
=
∂x1
∂x2
∂x2
⇒
∂φ(ψ1(x2, y), x2 , y) ∂x1(x2 , y)
∂φ(ψ1(x2 , y), x2, y)
= −
∂x1
∂x2
∂x2
⇒
(64)
− ∂φ(ψ1 (x2 , 2y), x2 , y)
∂x1(x2 , y)
= ∂φ(ψ (x ∂x
1 2 , y), x2 , y)
∂x2
∂x1
Consider the following example.
φ(x01, x02, y 0 ) = 0
(65)
y 0 − f (x01, x02 ) = 0
The theorem says that we can solve the equation for x01 .
x01 = ψ1(x02, y 0 )
(66)
It is also true that
φ(ψ1(x2, y), x2 , y) = 0
(67)
y − f (ψ1(x2, y), x2) = 0
Now compute the relevant derivatives
∂φ(ψ1(x2, y), x2 , y)
∂f (ψ1(x2, y), x2)
= −
∂x1
∂x1
∂φ(ψ1(x2, y), x2 , y)
∂f (ψ1(x2, y), x2)
= −
∂x2
∂x2
The theorem then says that
" ∂φ(ψ1 (x2 , y), x2 , y) #
∂x1(x2, y)
2
= − ∂φ(ψ (x∂x, y),
x2 , y)
1 2
∂x2
(68)
∂x1
= −
= −
"
2 , y),x2 )
− ∂f (ψ1(x
∂x2
#
2 , y),x2 )
− ∂f (ψ1 (x
∂x1
(69)
∂f (ψ1 (x2 , y),x2 )
∂x2
∂f (ψ1 (x2 , y),x2 )
∂x1
4.3. Example with two equations and three variables. Consider the following system of
equations
φ1(x1 , x2, y) = 3x1 + 2x2 + 4y = 0
φ2 (x1 , x2, y) = 4x1 + x2 + y = 0
(70)
22
GENERAL ANALYSIS OF MAXIMA/MINIMA IN CONSTRAINED OPTIMIZATION PROBLEMS
The Jacobian is given by
"
∂φ1
∂x1
∂φ2
∂x1
∂φ1
∂x2
∂φ2
∂x2
#
3 2
=
4 1
(71)
We can solve system 70 for x1 and x2 as functions of y. Move y to the right hand side in
each equation.
3x1 + 2x2 = −4y
(72a)
4x1 + x2 = −y
(72b)
Now solve equation 72b for x2
x2 = −y − 4x1
(73)
Substitute the solution to equation 73 into equation 72a and simplify
3x1 + 2(−y − 4x1) = −4y
⇒ 3x1 − 2y − 8x1 = −4y
⇒ −5x1 = −2y
(74)
2
y = ψ1(y)
5
Substitute the solution to equation 74 into equation 73 and simplify
⇒ x1 =
x2 = −y − 4
2
y
5
5
8
y − y
5
5
13
= −
y = ψ2(y)
5
If we substitute these expressions for x1 ad x2 into equation 70 we obtain
⇒ x2 = −
φ1
2
13
2
13
y , − y, y ) = 3 y + 2 − y + 4y
5
5
5
5
6
26
20
= y −
y +
y
5
5
5
20
20
= −
y +
y = 0
5
5
and
(75)
(76)
GENERAL ANALYSIS OF MAXIMA/MINIMA IN CONSTRAINED OPTIMIZATION PROBLEMS
φ2
23
2
13
2
13
y , − y, y ) = 4 y + − y + y
5
5
5
5
8
13
5
= y −
y + y
5
5
5
13
13
= y −
y = 0
5
5
(77)
Furthermore
∂ψ1)
2
=
∂y
5
∂ψ2)
13
= −
∂y
5
We can solve for these partial derivatives using equation 60 as follows
∂φ1 ∂ψ1
∂φ1 ∂ψ2
−∂φ1
+
=
∂x1 ∂y
∂x2 ∂y
∂y
∂φ2 ∂ψ1
∂φ2 ∂ψ2
−∂φ2
+
=
∂x1 ∂y
∂x2 ∂y
∂y
Now substitute in the derivatives of φ1 and φ2 with respect to x1 , x2 , and y.
Solve equation 80b for
(78)
(79a)
(79b)
3
∂ψ1
∂ψ2
+ 2
= −4
∂y
∂y
(80a)
4
∂ψ1
∂ψ2
+ 1
= −1
∂y
∂y
(80b)
∂ψ2
∂y
∂ψ2
∂ψ1
= −1 − 4
∂y
∂y
Now substitute the answer from equation 81 into equation 80a
∂ψ1
∂ψ1
3
+ 2 −1 − 4
= −4
∂y
∂y
⇒ 3
∂ψ1
∂ψ1
− 2 − 8
= −4
∂y
∂y
∂ψ1
⇒ −5
= −2
∂y
∂ψ1
2
=
∂y
5
If we substitute equation 82 into equation 81 we obtain
⇒
(81)
(82)
24
GENERAL ANALYSIS OF MAXIMA/MINIMA IN CONSTRAINED OPTIMIZATION PROBLEMS
∂ψ2
∂ψ1
= −1 − 4
∂y
∂y
∂ψ2
2
⇒
= −1 − 4
∂y
5
=
5. F ORMAL A NALYSIS
OF
(83)
−5
8
13
−
= −
5
5
5
L AGRANGIAN M ULTIPLIERS
P ROBLEMS
AND
E QUALITY C ONSTRAINED
5.1. Definition of the Lagrangian. Consider a function on n variables denoted
f (x) = f (x1, x2 , . . . , xn ). Suppose x∗ minimizes f (x) for all x Nδ (x∗ ) that satisfy
gi (x) = 0
i = 1, . . . , m
Assume the Jacobian matrix (J) of the constraint equations gi (x∗ ) has rank m. Then:
∇f (x∗ ) =
m
X
λ∗i ∇gi(x∗ )
(84)
i=1
In other words the gradient of f at x∗ is a linear combination of the gradients of gi at x∗
with weights λ∗i . For later reference note that the Jacobian can be written


∂g1(x∗) ∂g2(x∗ )
∂gm (x∗)
...
 ∂x1
∂x1
∂x1 


 ∂g (x∗) ∂g (x∗ )
∗) 
(x
∂g
2
m
 1

...


∂x2
∂x2 
 ∂x2
(85)
Jg = 



..
..
..
..


.
.
.
.




∗ 
 ∂g1(x∗) ∂g2(x∗ )
∂gm (x )
...
∂xn
∂xn
∂xn
Proof:
By suitable rearrangement of the rows
we can
always assume the m × m matrix formed
from the first m rows of the Jacobian
equations:
m
X
∂gi (x∗)
i=1
∂xj
∂gi (x∗ )
∂xj
λj =
is non-singular. Therefore the set of linear
∂f (x∗)
∂xj
j = 1, . . . , m
(86)
will have a unique solution λ∗. In matrix notation we can write equation 86 as
Jλ = ∇f
If J is invertible, we can solve the system for λ. Therefore (84) is true for the first m
elements of ∇f (x∗).
GENERAL ANALYSIS OF MAXIMA/MINIMA IN CONSTRAINED OPTIMIZATION PROBLEMS
25
We must show (84) is also true for the last n−m elements. Let x̃ = (xm+1 , xm+2 , . . . , xn ).
Then by using the implicit function theorem we can solve for the first m xs in terms of the
remaining xs or x̃ .
x∗j = hj (x̃∗)
j = 1, . . ., m
(87)
∗
We can define f (x ) as
f (x∗ ) = f (h1 (x̃∗), h2 (x̃∗) . . . hm (x̃∗ ), x∗m+1 . . . x∗n )
(88)
Since we are at a minimum, we know that the first partial derivatives of f with respect
to xm+1 , xm+2 , . . . , xn must vanish at x∗ , i.e.
∂f (x∗ )
=0
∂xj
j = m + 1, . . ., n
Totally differentiating (88) we obtain
m
∂f (x∗) X ∂f (x∗) ∂hk (x̃∗ ) ∂f (x∗)
=
+
=0
∂xj
∂xk
∂xj
∂xj
k=1
(89)
j = m + 1, . . . , n
by the implicit function theorem. We can also use the implicit function theorem to find the
derivative of the ith constraint with respect to the jth variable where the jth variable goes
from m + 1 to n. Applying the theorem to
gi (x∗) = gi (h1 (x̃∗), h2(x̃∗ ) . . . hm (x̃∗ ), x∗m+1 . . . x∗n ) = 0
we obtain
m
X
∂gi (x∗) ∂hk (x̃∗)
k=1
∂xk
∂xj
=
−∂gi(x∗ )
∂xj
i = 1, . . ., m
(90)
Now multiply each side of (90) by λ∗i and add them up.
m
m X
X
λ∗i
i=1 k=1
∂gi(x∗) ∂hk (x̃∗ )
∂gi(x∗ )
+ λ∗i
=0
∂xk
∂xj
∂xj
(91)
j = m + 1, . . . , n
Now subtract (91) from (89) to obtain:
"
#
m
m
m
X
∂f (x∗) X ∗ ∂gi(x∗)
∂f (x∗) X ∗ ∂gi(x∗ )
+
−
λi
−
λi
=0
∂xk
∂xk
∂xj
∂xj
k=1
i=1
i=1
j = m + 1, . . . , n
(92)
26
GENERAL ANALYSIS OF MAXIMA/MINIMA IN CONSTRAINED OPTIMIZATION PROBLEMS
The bracket term is zero from (86) so that
m
∂f (x∗) X ∗ ∂gi(x∗ )
−
λi
=0
∂xj
∂xj
j = m + 1, . . ., n
(93)
i=1
Since (86) implies this is true, for j = 1, . . . , m we know it is true for j = 1, 2, . . ., n and
we are finished.
The λi are called Lagrange multipliers and the expression
L(x, λ) = f (x) −
m
X
λi gi (x)
(94)
i=1
is called the Lagrangian function.
5.2. Proof of Necessary Conditions. The necessary conditions for an extreme point are
∇L(x∗, λ∗) = ∇f (x∗ ) − Jg (x∗ )λ = 0
(95)
m
⇒
∂f (x∗) X ∗ ∂gi (x∗)
−
λi
=0
∂xj
∂xj
j = m + 1, . . . , n
i=1
This is obvious from (84) and (94).
5.3. Proof of Sufficient Conditions. The sufficient conditions are repeated here for convenience
Let f, g1, . . . , gm be twice continuously differentiable real-valued functions on Rn . If
there exist vectors x∗ Rn , λ∗ Rm such that
∇L(x∗ λ) = 0
(5)
and for every non-zero vector z Rn satisfying
z 0 ∇gi (x∗) = 0, . . . i = 1, . . . , m
(6)
it follows that
z 0 ∇2x L(x∗, λ∗)z > 0
(7)
then f has a strict local minimum at x∗ , subject to gi(x) = 0, i = 1, . . . , , m. If the
inequality in (7) is reversed , then f has strict local maximum at x∗ .
Proof:
Assume x∗ is not a strict local minimum. Then there exists a neighborhood Nδ (x∗ ) and
a sequence {z k }, zk Nδ (x∗), z k 6= x∗ converging to x∗ such that for every zk {zk}.
gi (z k ) = 0
i = l, . . . , m
f (x∗) ≥ f (z k )
(96)
(97)
GENERAL ANALYSIS OF MAXIMA/MINIMA IN CONSTRAINED OPTIMIZATION PROBLEMS
27
This simply says that since x∗ is not the minimum value subject to the constraints there
exists a sequence of values in the neighborhood of x∗ that satisfies the constraints and has
an objective function value less than or equal to f (∗).
The proof will require the mean value theorem which is repeated here for completeness.
Mean Value Theorem
Theorem 2. Let f be defined on an open subset (Ω) of Rn and have values in R1 . Suppose the set
Ω contains the points a,b and the line segment S joining them, and that f is differentiable at every
point of this segment. Then there exists a point c on S such that
f (b) − f (a) = ∇f (c)0(b − a)
∂f (c)
∂f (c)
∂f (c)
(b1 − a1 ) +
(b2 − a2 ) + · · · +
(bn − an )
∂x1
∂x2
∂xn
where b is the vector (b1, b2 , . . ., bn ) and a is the vector (a1, a2 , . . . , an ).
=
(98)
Now let y k and z k be vectors in Rn and let z k = x∗ + θk y k where θk > 0 and || y k ||= 1
so that z k − x∗ = θk y k . The sequence {θk , y k } has a subsequence that converges to (0, ȳ)
where || y ||= 1. Now if we use the mean value theorem we obtain for each k in this
subsequence
gi(z k ) − gi (x∗) = θk y k0 ∇gi (x∗ + γik θk y k ) = 0, i = 1, . . . , m
(99)
k
where γi is a number between 0 and 1 and gi is the ith constraint. The expression is
equal to zero because we assume that the constraint is satisfied at the optimal point and at
the point z k by equation 98.
Expression 99 follows from the mean value theorem because z k − x∗ = θk y k and with
γik between zero and one, γik θk y k is between z k = x∗ + θk and x∗
If we use the mean value theorem to evaluate f (zk ) we obtain
f (z k ) − f (x∗) = θk y k0 ∇f (x∗ + η k θk y k ) ≤ 0
where 0 < ηk < 1. This is less than zero by our assumption in equation 97.
If we divide (99) and (100) by θk and take the limit as k → ∞ we obtain
h
i
lim y k0 ∇gi (x∗ + η k θk y k ) = ȳ 0 ∇gi(x∗ ) = 0 i = 1, 2, . . . , m
k→∞
h
i
lim y k0 ∇f (x∗ + η k θk y k ) = ȳ 0 ∇fi (x∗) ≤ 0
k→∞
(100)
(101)
(102)
Now remember from Taylor’s theorem that we can write the Lagrangian in (95) as
L(z k , λ∗) = L(x∗, λ∗) + (z k − x∗ )0∇x L(x∗, λ∗)
1 2
+ θk (z k − x∗)0 ∇2x L(x∗ + β k θk y k , λ∗)(z k − x∗ )
2
1 2
= L(x∗, λ∗) + θk y k0∇x L(x∗, λ∗ ) + θk y k0∇2x L(x∗ + β k θk y k , λ∗)y k
2
(103)
28
GENERAL ANALYSIS OF MAXIMA/MINIMA IN CONSTRAINED OPTIMIZATION PROBLEMS
where 0 < β k < 1.
Now note that
L(z k , λ∗ ) = f (z k ) −
m
X
L(x∗ , λ∗ ) = f (x∗) −
m
X
λigi (z k )
i=1
λigi (x∗ )
i=1
and that at the optimum or at the assumed point z k , gi (·) = 0.
Also ∇L(x∗ , λ∗ ) = 0 at the optimum so the second term on the right hand side of (103)
is zero. Move the first term to the left hand side to obtain
1 k2 k0 2
(104)
θ y ∇x L(x∗ + β k θk y k , λ∗)y k
2
Because we assumed f (x∗) ≥ f (z k ) in (97) and that g(·) is zero at either x∗ or z k , it is
clear that
L(z k , λ∗) − L(x∗, λ∗) =
L(z k , λ∗) − L(x∗ , λ∗ ) ≤ 0
(105)
Therefore,
1 k2 k0 2
θ y ∇xL(x∗ + β k θk y k , λ∗)y k ≤ 0
2
2
Divide both sides by 12 θk to obtain
0
(106)
y k ∇2x L(x∗ + β k θk y k , λ∗)y k ≤ 0
Now take the limit as k → ∞ to obtain
(107)
ȳ 0 ∇2x L(x∗, λ∗)ȳ ≤ 0
We are finished since ȳ =
6 0, and by equation 101,
(108)
ȳ 0 ∇gi(x∗ ) = 0,
i = 1, 2, . . . , m
that is, if x∗ is not a minimum then we have a non-zero vector y satisfying
ȳ 0 ∇gi(x∗ ) = 0, i = 1, 2, . . . , m
(109)
0
2
∗
∗
∗
where ȳ ∇x L(x , λ )ȳ ≤ 0 . But if x is a minimum then equation 6 rather than (108)
will hold.